My textbook writes the following proof for the statement in the title:

For a homomorphism $\varphi:G \to G'$ and any $g\in G$, $\varphi(g) = \varphi(ge) = \varphi(g)\varphi(e)$. Multiplying both sides of this equation on the left by $(\varphi(g))^{−1}$ gives the desired result, $e′ = e′\varphi(e) = \varphi(e)$.

It's not clear to me why we should be able to work like this. In particular, we only have that $\varphi(g) = \varphi(g)\varphi(e)$ for all elements in $G'$ of the form $\varphi(g)$. But it seems like the proof is tacitly using that $e'$ is the unique element for which $e'g' = g'$ for all $g \in G'$...and yet we haven't shown that $\varphi(e)$ actually does this for every $g'$!

  • $\begingroup$ No, this is not tacitly assumed. We just take the equation $\phi(g)=\phi(g)\phi(e)$ and cancel the invertible element $\phi(g)$. But $\phi(g)\cdot \phi(g)^{-1}=1$ in $G'$, and the $1$ is the $e'$. We obtain $e'=1=\phi(e)$. Take for example $G=G'=\Bbb Z$ and $\phi(n)=2n$. $\endgroup$ Mar 19 at 20:21
  • $\begingroup$ Use $\varphi$ for $\varphi$ and $\in$ for $\in$. $\endgroup$
    – Shaun
    Mar 19 at 20:23
  • $\begingroup$ @DietrichBurde I guess I am looking for an argument which doesn't just say "cancel" but shows what that means. For example, suppose we have the lemma that in a group $G$, if $gh = gk$ then $h=k$.. Computing, we have $h = eh = (g^{-1}g)h = g^{-1}(gh) = g^{-1}(gk) = (g^{-1}g) k = ek = k. $ I was trying to convince myself using this more direct line. $\endgroup$
    – EE18
    Mar 19 at 20:25
  • $\begingroup$ Cancelling in a group really means $g\cdot g^{-1}=e$, so $gh=gh'$ gives $h=h'$. This is all. So it is just the existence of $g^{-1}$. The cancelling in the multiplicative group of a field, say, of real numbers is the same. If $2a=2b$, then $a=b$, because $2$ has an inverse $1/2$, which we can multiply from the left. $\endgroup$ Mar 19 at 20:26
  • $\begingroup$ In a group, if $xy=x$ for a single $x$, then $y=e$. $\endgroup$ Mar 19 at 20:29

1 Answer 1


Because every element in a group has an inverse, it's enough for

$$fg' = g'$$

for just one $g'\in G'$ to prove that $f$ is the identity.

The fact that we already know that $G'$ is a group has done most of the work for us - we know that there is an element in $G'$ that preserves every element when you multiply by it, we just have to show that that's what $\varphi$ maps $e$ to.

  • $\begingroup$ Ah I see. I think I can use the lemma I proved in the comments above to show this? $\endgroup$
    – EE18
    Mar 19 at 20:27
  • 1
    $\begingroup$ Yup, that lemma is a great proof. And it's exactly what mathematicians mean when they say "cancel" in a group setting. :-) $\endgroup$
    – JonathanZ
    Mar 19 at 20:31
  • $\begingroup$ Awesome, thank you! $\endgroup$
    – EE18
    Mar 19 at 20:38
  • $\begingroup$ As an aside, does this mean that if we relax the "for any $g \in G$" and talk about a map such that for exactly one $g \in G$ we have $f(g) = f(ge) = f(g)f(e)$ then $f$ preserves identities? $\endgroup$
    – EE18
    Mar 19 at 20:57
  • $\begingroup$ Hmm, I don't think I understand what you are asking., especially with the "exactly one $g$" specification. $f(g) = f(ge)$ automatically for all $g \in G$, not just one. - that's $e$ being the identity in $G$. And $f(ge) = f(g)f(e)$ is also true for all $g$ - that's what being a homomorphism is. If you're asking if we can have weaker assumptions than having a group identity or the homomorphism rule, and still deduce preserving identities, then maybe you should check out monoids or semi-groups and see how things that work in groups can stop working in those structures. $\endgroup$
    – JonathanZ
    Mar 19 at 21:24

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